# Solve this

Question:

Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$. Then the

vector product $(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times((\overrightarrow{\mathrm{a}} \times((\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{b}})) \times \overrightarrow{\mathrm{b}})$ is

equal to:

1. $5(34 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$

2. $7(34 \hat{i}-5 \hat{j}+3 \hat{k})$

3. $7(30 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+7 \hat{\mathrm{k}})$

4. $5(30 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+7 \hat{\mathrm{k}})$

Correct Option: 2,

Solution:

$\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{b}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

$\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}=3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}} ; \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=-1+2+6=7$

$((\vec{a} \times((\vec{a}-\vec{b}) \times \vec{b})) \times \vec{b})$

$((\vec{a} \times(\vec{a} \times \vec{b}-\vec{b} \times \vec{b})) \times \vec{b})$

$(\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}-0)) \times \overrightarrow{\mathrm{b}}$

$((\vec{a} \cdot \vec{b}) \vec{a}-(\vec{a} \cdot \vec{a}) \vec{b}) \times \vec{b}$

$(\vec{a} \cdot \vec{b}) \vec{a} \times \vec{b}-(\vec{a} \cdot \vec{a})(\vec{b} \times \vec{b})$

$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & 1 & 2 \\ -1 & 2 & 3\end{array}\right|=-\hat{\mathrm{i}}-5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$

$\therefore 7(-\hat{\mathbf{i}}-\mathbf{5} \hat{\mathbf{j}}+3 \hat{\mathbf{k}})$

$(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}) \times(7(-\hat{\mathrm{i}}-5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}))$

$7(0 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}) \times(-\hat{\mathbf{i}}-5 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})$

$\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & 3 & 5 \\ -1 & -5 & 3\end{array}\right|$

$\Rightarrow 34 \hat{\mathbf{i}}-(5) \hat{\mathbf{j}}+(3 \hat{\mathbf{k}})$

$\Rightarrow 34 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$

$\therefore \mathbf{7}(34 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})$